Question

Let \( y = y(x) \) be the solution of the differential equation \[\left( 1 + x^2 \right) \frac{dy}{dx} + y = e^{\tan^{-1}x}, \, y(1) = 0.\]Then \( y(0) \) is:

• A. \( \frac{1}{4} \left( e^{\pi/2} - 1 \right) \)
• B. \( \frac{1}{2} \left( 1 - e^{\pi/2} \right) \)
• C. \( \frac{1}{4} \left( 1 - e^{\pi/2} \right) \)
• D. \( \frac{1}{2} \left( e^{\pi/2} - 1 \right) \)

Answer 1

The Correct Option is B

To solve the given differential equation, we notice that it is a first-order linear differential equation of the form:

\[(1 + x^2) \frac{dy}{dx} + y = e^{\tan^{-1}x}.\]

We need to make this equation into the standard form:

\[\frac{dy}{dx} + \frac{y}{1 + x^2} = \frac{e^{\tan^{-1}x}}{1 + x^2}.\]

The integrating factor (IF) for this equation is given by:

\[IF = e^{\int \frac{1}{1+x^2} \, dx}= e^{\tan^{-1}x}.\]

Multiplying the entire differential equation by this integrating factor, we have:

\[e^{\tan^{-1}x}\left(\frac{dy}{dx} + \frac{y}{1 + x^2}\right) = \frac{e^{2\tan^{-1}x}}{1 + x^2}.\]

This simplifies to:

\[\frac{d}{dx}\left(ye^{\tan^{-1}x}\right) = \frac{e^{2\tan^{-1}x}}{1 + x^2}.\]

Integrating both sides, we have:

\[ye^{\tan^{-1}x} = \int \frac{e^{2\tan^{-1}x}}{1 + x^2} \, dx + C.\]

Now, to evaluate the integral, let \(t = \tan^{-1}x\), which implies \(\frac{dt}{dx} = \frac{1}{1+x^2}\) or \(dx = (1 + x^2) \, dt\). Thus, the integral becomes:

\[\int e^{2t} \, dt = \frac{e^{2t}}{2} + C.\]

Substituting back, we have:

\[ye^{\tan^{-1}x} = \frac{e^{2\tan^{-1}x}}{2} + C.\]

Substituting the initial condition \(y(1) = 0\) when \(x = 1\):

\[0 = \frac{e^{\pi/4}}{2} + C.\]

This gives us:

\[C = -\frac{e^{\pi/2}}{2}.\]

Substituting C back into the solution, we have:

\[ye^{\tan^{-1}x} = \frac{e^{2\tan^{-1}x}}{2} - \frac{e^{\pi/2}}{2}.\]

Therefore, solving for \(y\), we have:

\[y = \frac{e^{\tan^{-1}x}}{2} - \frac{e^{\tan^{-1}x-\pi/2}}{2}.\]

To find \(y(0)\), substitute \(x = 0\), which gives:

\[y(0) = \frac{1}{2} - \frac{1}{2} e^{\pi/2}.\]

Thus, the value of \(y(0)\) is:

\( \frac{1}{2} \left( 1 - e^{\pi/2} \right) \)

Answer 2

The given differential equation is:

\[ \frac{dy}{dx} + \frac{y}{1+x^2} = \frac{e^{\tan^{-1}x}}{1+x^2}. \]

Integrating factor (I.F.):

\[ \text{I.F.} = e^{\int \frac{1}{1+x^2} dx} = e^{\tan^{-1}x}. \]

Multiply through by I.F.:

\[ y \cdot e^{\tan^{-1}x} = \int e^{\tan^{-1}x} \cdot \frac{e^{\tan^{-1}x}}{1+x^2} dx + C. \]

Simplify:

\[ y \cdot e^{\tan^{-1}x} = \int e^{2\tan^{-1}x} \cdot \frac{1}{1+x^2} dx + C. \]

Substitute:

\[ \tan^{-1}x = z, \quad \frac{1}{1+x^2} dx = dz. \]

Thus:

\[ y \cdot e^{\tan^{-1}x} = \frac{e^{2z}}{2} + C. \]

Apply initial condition: \( y(1) = 0 \), \( \tan^{-1}(1) = \frac{\pi}{4} \):

\[ 0 = \frac{e^{\pi/2}}{2} + C \implies C = -\frac{e^{\pi/2}}{2}. \]

Finally:

\[ y \cdot e^{\tan^{-1}x} = \frac{e^{2\tan^{-1}x}}{2} - \frac{e^{\pi/2}}{2}. \]

At \( x = 0 \), \( \tan^{-1}(0) = 0 \):

\[ y(0) = \frac{1}{2} \left( 1 - e^{-\pi/2} \right). \]